3 DERIVATIVES DERIVATIVES The functions that we have

3 DERIVATIVES

DERIVATIVES The functions that we have met so far can be described by expressing one variable explicitly in terms of another variable. § For example, or in general y = f(x). , or y = x sin x,

DERIVATIVES However, some functions are defined implicitly.

DERIVATIVES 3. 6 Implicit Differentiation In this section, we will learn: How functions are defined implicitly.

IMPLICIT DIFFERENTIATION Equations 1 and 2 Some examples of implicit functions are: x 2 + y 2 = 25 x 3 + y 3 = 6 xy

IMPLICIT DIFFERENTIATION In some cases, it is possible to solve such an equation for y as an explicit function (or several functions) of x. § For instance, if we solve Equation 1 for y, we get § So, two of the functions determined by the implicit Equation 1 are and

IMPLICIT DIFFERENTIATION The graphs of f and g are the upper and lower semicircles of the circle x 2 + y 2 = 25.

IMPLICIT DIFFERENTIATION It’s not easy to solve Equation 2 for y explicitly as a function of x by hand. § A computer algebra system has no trouble. § However, the expressions it obtains are very complicated.

FOLIUM OF DESCARTES Nonetheless, Equation 2 is the equation of a curve called the folium of Descartes shown here and it implicitly defines y as several functions of x.

FOLIUM OF DESCARTES The graphs of three functions defined by the folium of Descartes are shown.

IMPLICIT DIFFERENTIATION When we say that f is a function defined implicitly by Equation 2, we mean that the equation x 3 + [f(x)]3 = 6 x f(x) is true for all values of x in the domain of f.

IMPLICIT DIFFERENTIATION Fortunately, we don’t need to solve an equation for y in terms of x to find the derivative of y.

IMPLICIT DIFFERENTIATION METHOD Instead, we can use the method of implicit differentiation. § This consists of differentiating both sides of the equation with respect to x and then solving the resulting equation for y’.

IMPLICIT DIFFERENTIATION METHOD In the examples, it is always assumed that the given equation determines y implicitly as a differentiable function of x so that the method of implicit differentiation can be applied.

IMPLICIT DIFFERENTIATION a. If x 2 + y 2 = 25, find Example 1 . b. Find an equation of the tangent to the circle x 2 + y 2 = 25 at the point (3, 4).

IMPLICIT DIFFERENTIATION Example 1 a Differentiate both sides of the equation x 2 + y 2 = 25:

IMPLICIT DIFFERENTIATION Example 1 a Remembering that y is a function of x and using the Chain Rule, we have: Then, we solve this equation for :

IMPLICIT DIFFERENTIATION E. g. 1 b—Solution 1 At the point (3, 4) we have x = 3 and y = 4. So, § Thus, an equation of the tangent to the circle at (3, 4) is: y – 4 = – ¾(x – 3) or 3 x + 4 y = 25.

IMPLICIT DIFFERENTIATION E. g. 1 b—Solution 2 Solving the equation x 2 + y 2 = 25, we get: § The point (3, 4) lies on the upper semicircle § So, we consider the function

IMPLICIT DIFFERENTIATION E. g. 1 b—Solution 2 Differentiating f using the Chain Rule, we have:

IMPLICIT DIFFERENTIATION E. g. 1 b—Solution 2 So, § As in Solution 1, an equation of the tangent is 3 x + 4 y = 25.

NOTE 1 The expression dy/dx = -x/y in Solution 1 gives the derivative in terms of both x and y. It is correct no matter which function y is determined by the given equation.

NOTE 1 For instance, for , we have: However, for we have: ,

IMPLICIT DIFFERENTIATION Example 2 a. Find y’ if x 3 + y 3 = 6 xy. b. Find the tangent to the folium of Descartes x 3 + y 3 = 6 xy at the point (3, 3). c. At what points in the first quadrant is the tangent line horizontal?

IMPLICIT DIFFERENTIATION Example 2 a Differentiating both sides of x 3 + y 3 = 6 xy with respect to x, regarding y as a function of x, and using the Chain Rule on y 3 and the Product Rule on 6 xy, we get: 3 x 2 + 3 y 2 y’ = 6 xy’ + 6 y or x 2 + y 2 y’ = 2 xy’ + 2 y

IMPLICIT DIFFERENTIATION Now, we solve for y’: Example 2 a

IMPLICIT DIFFERENTIATION When x = y = 3, § A glance at the figure confirms that this is a reasonable value for the slope at (3, 3). § So, an equation of the tangent to the folium at (3, 3) is: y – 3 = – 1(x – 3) or x + y = 6. Example 2 b

IMPLICIT DIFFERENTIATION Example 2 c The tangent line is horizontal if y’ = 0. § Using the expression for y’ from (a), we see that y’ = 0 when 2 y – x 2 = 0 (provided that y 2 – 2 x ≠ 0). § Substituting y = ½x 2 in the equation of the curve, we get x 3 + (½x 2)3 = 6 x(½x 2) which simplifies to x 6 = 16 x 3.

IMPLICIT DIFFERENTIATION Example 2 c Since x ≠ 0 in the first quadrant, we have x 3 = 16. If x = 161/3 = 24/3, then y = ½(28/3) = 25/3.

IMPLICIT DIFFERENTIATION Example 2 c Thus, the tangent is horizontal at (0, 0) and at (24/3, 25/3), which is approximately (2. 5198, 3. 1748). § Looking at the figure, we see that our answer is reasonable.

NOTE 2 There is a formula for the three roots of a cubic equation that is like the quadratic formula, but much more complicated.

NOTE 2 If we use this formula (or a computer algebra system) to solve the equation x 3 + y 3 = 6 xy for y in terms of x, we get three functions determined by the following equation.

NOTE 2 and

NOTE 2 These are three functions whose graphs are shown in the earlier figure.

NOTE 2 You can see that the method of implicit differentiation saves an enormous amount of work in cases such as this.

NOTE 2 Moreover, implicit differentiation works just as easily for equations such as y 5 + 3 x 2 y 2 + 5 x 4 = 12 for which it is impossible to find a similar expression for y in terms of x.

IMPLICIT DIFFERENTIATION Example 3 Find y’ if sin(x + y) = y 2 cos x. § Differentiating implicitly with respect to x and remembering that y is a function of x, we get: § Note that we have used the Chain Rule on the left side and the Product Rule and Chain Rule on the right side.

IMPLICIT DIFFERENTIATION Example 3 If we collect the terms that involve y’, we get: So,

IMPLICIT DIFFERENTIATION Example 3 The figure, drawn with the implicit-plotting command of a computer algebra system, shows part of the curve sin(x + y) = y 2 cos x. § As a check on our calculation, notice that y’ = -1 when x = y = 0 and it appears that the slope is approximately -1 at the origin.

IMPLICIT DIFFERENTIATION The following example shows how to find the second derivative of a function that is defined implicitly.

IMPLICIT DIFFERENTIATION Example 4 Find y” if x 4 + y 4 = 16. § Differentiating the equation implicitly with respect to x, we get 4 x 3 + 4 y 3 y’ = 0.

IMPLICIT DIFFERENTIATION Solving for y’ gives: E. g. 4—Equation 3

IMPLICIT DIFFERENTIATION Example 4 To find y’’, we differentiate this expression for y’ using the Quotient Rule and remembering that y is a function of x:

IMPLICIT DIFFERENTIATION Example 4 If we now substitute Equation 3 into this expression, we get:

IMPLICIT DIFFERENTIATION Example 4 However, the values of x and y must satisfy the original equation x 4 + y 4 = 16. So, the answer simplifies to: